text:
What practical alternatives are there to 100LL in 2021?
As a prospective student pilot in Southern California, I would like to avoid 100LL. Since the FAA 100LL phase-out is stalled, what other alternatives are there? Continental has a line of Diesel engines such as the CD-135, are supplemental type certificates available / retrofits common? What about "mogas?" Does the answer change if someone is looking to purchase an aircraft?

Of all the things to look for in a flight school, the type of fuel that their planes use seems like a very unusual priority. (OK, it'd be pretty cool if your first solo was in something that burns Jet-A... but you pretty much have to go the military route to make that happen.)

Closely related

Right now, Mogas is the only alternative fuel because the unleaded avgas program still seems to be going nowhere, although the industry testing program is supposed to restart this year.
You could put an aircraft diesel engine in your plane if you're rich, and run it on Jet A, but if you can afford that, you can probably afford other alternatives as well.
The most practical thing for a prospective owner to do is to buy an airplane and incorporate the Petersen Mogas STC (it's just a paperwork/placard mod for a lot of high wing airplanes, but physical parts may be required in other cases), and run the plane on mogas when possible, and avgas when not.
Overall, Mogas is better, because the tetraethyl lead in avgas used to get its octane rating is just bad news all around (the lead combustion byproducts contaminate the oil, leave crud in the cylinders, etc.) Mogas's main caveats are higher Reid Vapor Pressure (it boils more readily) and long term storage.  Mogas is more volatile and needs fuel stabilizers if a fuel supply is left for more than few months.  The volatility also means more susceptibility to vapour lock.
On airplanes with gravity feed fuel systems, vapour lock is less of a problem because there is positive fuel head pressure all the way from the tank to carb.  On low-wing pumped-fuel airplanes, there is pump inlet suction, which to the gas is like going up a several thousand feet depending on how hard the pump sucks, so on a very hot day with a hot engine, there is a chance of vapour lock and it can be prudent to install cooling shrouds on fuel system components like engine driven pumps (you'll know it's happening as the fuel pressure fluctuates).  I was having incipient vapour lock indications on my plane during heat waves (only on the ground) and installing a cooling shroud for my engine-driven fuel pump seems to have solved it.
Those issues aside, any 80 octane engine will be MUCH happier running on Mogas than 100LL, so if you can do it, it's a no-brainer.

Depending on the particular airplane, the Petersen STC may not be just paperwork.  On my Cherokee 180, it required replacing the OM fuel pump with dual pumps, along with some hoses and switches on the panel.  (Probably to deal with those vapor lock problems.)  Still, not a major project - IIRC (it was done over 20 years ago) it took about half a day.

Good information thanks. I have a homebuilt so didn't have to do an STC in the first place to use Mogas.

As of summer 2021, GAMI has been issued a STC for their unleaded G100UL which is to be a drop in replacement for 100LL. It is still unclear what the transition period will look like


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What descent rate would be a "hard" landing in a Huey or similar helicopter with skids?
I would like to know what vertical descent rate a Huey or other helicopter with skids can land safely at.  What would be considered a "hard" landing that would require air frame inspection?  What would the descent rate be for a running landing?  Thanks.

You won't get a hard number and will only get opinions.  In the end, a landing that doesn't bend anything permanently is fine.  If you touchdown unusually hard, it'll be an airmanship issue for the pilot to decide what is "hard" and whether or not an inspection of the skids or some other structure is appropriate.  I'd probably get out and look for bent skids with a touchdown more than 100 or 200 fpm, or whatever just felt rough.  If you really do it hard, you will either collapse the skids, or chop the tail boom off and it won't require any judgement.

The manufacturer should have some specifications on how hard a landing the gear can take, although that's also dependent on weight. A 100fpm descent with a full load is more force on the gear than a light load.

@JohnK the airframe is designed and manufactured to an ultimate descent rate, above which there will be deformations.

A particular issue is a hard landing in which one skid or wheel touches the ground first and the rotor has drag hinges. This can cause the blades to bunch up on one side of the rotor so it develops an asymmetry and starts to wobble. If the frequency of this wobble resonates with the eigenfrequency of the landing gear, ground resonance will develop. What follows can be watched here.
I would be less concerned about the sink rate and more about the symmetry of ground contact. To give only a single number is not sufficient for characterizing a safe landing.

I don't have descent rate numbers, but many years ago I trained on a Bell 206. In the engine/transmission compartment, I recall there being a spike which was attached to the drivetrain, the drive train having some freedom of movement within the compartment. As the drivetrain wiggled around, this spike would move with it, and there was a plate with an opening within which the spike moved. If there was a hard landing, the spike could contact the sides of the opening. This could leave a mark, deform and in extreme cases could even break off. I Googled and found a manual for the 206, it mentions the skids and crosstubes will deform at 2.5G of load. It also mentions if the cabin fuselage reaches the ground, it is considered to have sustained a 10G load.
Link to the document:
https://www.bellcustomer.com/Bulletins/Download?FileName=206AB-Inspection_and_Airworthiness_Limitations.pdf&CategoryID=139
Neat, I wouldn't have thought the manual for the 205 (aka Huey for the military version UH-1) was also available but it is:
See page 95 for the information on hard landing. No specific descent rates mentioned, but rather procedures to check for specific types of damage are spelled out.
https://www.bellcustomer.com/Bulletins/Download?filename=205A1-Inspection_and_Airworthiness_Limitations.pdf

A descent rate of over 240'/min on landing is considered hard.  It may damage the skids and is supposed to prompt an inspection of the skids before further use.
Touching down with a descent rate above 720'/min (at 16,700lb GW) risks damaging the airframe as well as the skids.

The precision in this answer is good, but could you include references as well? Possibly those values only apply to certain helicopters (such as the Huey), but I could see an argument that they're generic to all skids.

In an EC135T1 you can "land" with around 1000ft/min and you should survive it according to Airbus. So this is not an opinion but a fact.
Source: Crash Safety Documents EC135T1 Airbus

could you provide a link to said document? or is it not public?

source: Stack Exchange

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sort: votes


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ARINC429 - Encoding and Decoding of BINARY DATA
In ARINC429 protocol, how to encode and decode FAN DISCHARGE STATIC PRESSURE (LABEL - 127).
Range is 1.5 to 30.0
No of Sig. Bits - 11
Resolution is 0.016
What should be the value to be loaded into the word if my pressure value is 1.5 PSIA ?
As per my calculation, value need to be loaded = 1.5/0.016 = 93.75 which is a fractional value that cannot be loaded. So the value need to be scaled by some value.
Total no of data bits = 19 (including sign bit)
No of bits for representing value = 18
No of Significant bits are 11
Value need to be shifted by (18-11) which is equal to 7 bits => Max value = 128. So scaling factor will be 127 now.
Value before scaling = 93.75
Value after scaling = (93.75 x 128) = 12,000
Value to be loaded (bits-28 to 11) = 00 0010 1110 1110 0000b
But at the decoding side, the value will be decoded as 00 0010 1110 1b (i.e., Only 11 bits of data will be considered for decoding, and rest of the bits will be padding bits).
Decoded value = 00 0010 1110 1b = 93
Decoded Pressure value in PSIA = 93 x 0.016 = 1.488.
Please let me know, encoding and decoding is correct or not ?
Is it OK to take approximate value instead of actual value after decoding ?
What is the use of padding bits. Lets say in this example, i am using only 11 bits for representing a value due to which fractional part is getting discarded. So can i use these padding bits to represent fractional value ? In that case, How the value gets decoded ?
Thanks in advance.

Going off of the resolution, it looks to me like the actual label range would be 0-32.0, which with 11 significant bits gives an LSB of $\frac{32}{2048} = \frac{1}{64} = 0.015625$.
$1.5 * 64 = 96$

Thanks for answering my question.

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What do we mean by taking the lowest of ZFW; TOW and LW to determine the maximum takeoff weight of the day?
Firstly, I understand that the maximum permissible takeoff weight can be limited by landing weight because landing weight plus trip fuel gives us one take off weight. We obviously do not want to be taking off with any extra fuel which we can not burn to allow us to be under the MLW upon reaching the destination.
Secondly, I understand there is a structural limitation which can not be exceeded.
I can not get my head around how we are limited by the Zero fuel weight to determine how much should be the maximum takeoff weight for a given day?
I am specifically asking with regards to Airbus 320 wherein MTOW is 77t, MLW is 65t and MZFW is 62.5t

If you know how much fuel you'll need for the flight, then the things that can limit your load (which is really what you're interested in -- how much weight can I put on the airplane today, in terms of passengers + cargo) are:

Empty weight + Payload (ZFW) may not exceed Max ZFW
ZFW + fuel for the flight (TOW) may not exceed MTOW
TOW - fuel burned during flight (i.e. ZFW + fuel reserves) may not exceed MLW

Depending on how much fuel you expect to burn and how much fuel you expect to land with (i.e. reserves), any of these may be the most limiting.
Max ZFW will be limiting for a short flight with minimal reserves required... you're taking off with minimal fuel and landing with minimal fuel, but if you loaded up to max TOW or max LW you'd be well over MZFW.
Max TOW will be limiting with lots of fuel burn planned: if you loaded the pax/cargo to MZFW, the fuel load would put you over MTOW.
Max Landing Weight tends to be most limiting when you need lots of fuel at landing for reserves.  (If you envision an extreme case where you need 5 hours of fuel still in the tanks at landing, if you were close to MZFW you'd be way, way over max landing weight, so the latter rather than the former is most limiting.)
Modern airliners tend to have a range of landing fuel that tends to suggest when MTOW or MLW will be the more limiting.  If you're planning to land with min reserves, you can load to MZFW easily enough (max ZFW + min reserve gas will be < max landing weight); you'll probably be limited by either MZFW or MTOW.  If you need to land with significant reserves, MLW will probably be most limiting for how much cargo/pax you can have.
If you can find a typical empty A-320 weight and typical "good weather" and "bad weather" fuel reserve values (maybe 3t and 7t for the latter), as well as some "short haul" and "long haul" burns (guessing, maybe 3t short haul and 20-25t long haul), you can work out various cases of what load you can accept in each scenario of good/bad wx and short/long haul flight using the equations I described above.
EDIT: The discussion above treats MTOW as a constant, although in reality that isn't necessarily the case: the max takeoff weight on a high, hot, and short runway can be significantly less than the structural max takeoff weight.  The basic math doesn't change, but it's fair to note that MTOW is the lesser of structural MTOW or the performance-limited takeoff weight.  (The various performance limits can be any of a number of things: accelerate-stop, accelerate-go, climb performance, etc.)  For some jets, you can also have a performance limitation of Enroute performance, i.e. being able to clear enroute terrain if an engine fails.  Most modern jets, however, have sufficient performance that this isn't often (or ever) an issue.  But it is the stuff of ground-eval orals.

Zero fuel weight has to do with wing flex limitations, it is the weight of everything except the fuel, which is contained in the wings as well as the center (fuselage) tanks. If the zero fuel weight is too high, once the fuel is used, the wing flex might put a lot of strain on the spars. Imagine a model airplane being suspended by strings through the center of lift of each wing. Then you add a lot of weight to the fuselage. You will put a lot of strain on the wing spars. The reason the center tank is taken into account is because of fuel distribution. I worked on freighters for Sterling Airways, now defunct, and a clever person came up with the idea of adding 2000KG of freight and always keeping the center tank 2000KG below the wing tanks. This is not normally allowed,but it was approved by the Danish CAA at the time.
Therefore, upon considering your first two issues, structural, landing weight and takeoff performance, as well as  ZFW + fuel reserves, the most limiting of the four is considered the maximum takeoff weight.

This is a good answer as to why ZFW is itself limited, along with takeoff weight & landing weight. +1


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Airbus 320 VS mode
I would like to understand how does the thrust control the Speed when we chose to fly in VS mode for climb or descent? I understand the priority is given to the rate of climb/descent when choosing to fly using the VS mode and the thrust always goes in the "Speed" mode but does it mean that the managed/selected speed will be maintained by the Auto thrust? I have observed that the thrust Speed mode causes variation on the speed target and may not necessarily maintain the speed while keeping the VS mode.
In the Climb phase:
When we pull the VS mode for climb, thrust goes in Speed, but what is the thrust value?
In the descent phase:
When we pull the VS mode for descent, thrust goes in Speed again, What if I want to maintain ROD 1500/min and reduce speed from 280kts to 250 kts at the same time? Will the Auto pilot guidance be able to do that?
Can anyone share more insights on this topic?

In speed mode, the autothrust varies engine thrust to maintain the pilot selected or the FMGC (Flight management guidance computer) calculated speed. In V/S mode, the priority is the selected vertical speed. If the vertical speed is high to the point, the set speed cannot be maintained with the set thrust, the aircraft will decelerate or accelerate depending on the circumstances.

In the Climb phase:
When we pull the VS mode for climb, thrust goes in Speed, but what is the thrust value?

That depends. If you are say maintaining 290 knots and you pull for V/S, vertical speed will go to whatever vertical speed that maintains 290 knots. At this point, if you increase your climb rate, then the aircraft will pitch up to maintain the target V/S. Thrust will also be added to maintain 290 knots. But once the thrust is limited (in a climb, max thrust is climb thrust), the V/S will be maintained at the expense of speed. i.e. the aircraft can no longer maintain 290 knots.

In the descent phase:
When we pull the VS mode for descent, thrust goes in Speed again, What if I want to maintain ROD 1500/min and reduce speed from 280kts to 250 kts at the same time? Will the Auto pilot guidance be able to do that?

In this situation, if you pull for a V/S of 1500 ft/ min, the aircraft will pitch down to maintain 1500 ft/ min. So, naturally, the thrust will go towards idle. Now, if you reduce speed, the thrust will go down again maybe to full idle in order to reduce speed. If the aircraft is at a V/S 1500 ft/ min when the speed is reduced and if the speed goes below 250 knots, then thrust will be added until 250 knots is maintained. If the aircraft is unable to maintain 250 knots with a V/S of 1500 ft/ min at idle thrust, then the thrust will remain at idle but the speed will increase to a point where it maintains 1500 ft/ min descent rate, again at the expense of speed.

Thanks, Anas Maaz for your quick prompt. Just to take my understanding a bit further, so as long as I have some amount of thrust still available on engines (above idle) or room for maximum thurst (in climb), the aircraft will try to attain the speed while keeping the target vertical speed? So if I am already at idle thurst with VS -1500 fpm and I try to reduce the speed further down, the speed will not be reduced? Will using the speed brakes with vertical speed mode be of any help to force the speed to reduce?


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What is the main and tail rotor speed (RPM) for a Sikorsky S52?
I need the RPMs for the main and tail rotor of a Sikorsky S52 for the research I am doing for my degree. I tried checking flight manuals and other websites, but I don't find concrete numbers, only percentage values.

It's variable depending on engine speed... Do you want the range? Should be easy to figure out if you have the gearbox ratios.

I can't find anything after a quick google,  but if you reached out to a museum with your question you may be able get a good answer, old guys like to talk about stuff like this, just a matter of getting the right person.

Given the $5m$ main rotor radius and the typical blade's tip speed due to the rotation around the hub of some $200m/s$, this gives:
$ \omega =  v_{tip} / r = 200/5 = 40 [rad/s] = 380rpm $
For the tail rotor this is proportionally bigger and equals $1900rpm$.
It could be that due to the technological limitations of those times, the $\omega$ was actually a bit lower but in absence of concrete numbers, these should be reasonable values to start with.

Wikipedia states the max speed as 180 km/h, which should be included in the tip speed calculation.

The OP asks for a definit source, not an assumption

I updated the answer with the previous comments


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How do the shockwaves (supersonic and hypersonic) interact with the structure?
My main question is why spaceplanes are conceptualized as pointy and lengthy, dart-like machines. What would happen if they were more like flying wing?
I'm hypothesizing a bit and I am not sure what would happen if a flying wing type of aircraft (most notable example B-2) were to go supersonic or hypersonic. So let's push B-2 to Mach 20.
How would the shockwaves interact with the structure? Would they remain oblique within supersonic regime (a)? Would they transform into bow shockwaves in hypersonic or cross the structure (b)?

and you've figured out why the shape of supersonic aircraft tends to be different from that of subsonic aircraft.

Aircraft with Flying wing configuration, such as the B2 bomber, fly at high subsonic speeds. During such a case, the airflow is at a high subsonic speed at the leading edge, and accerelates through to supersonic speeds as it moves over the front - upper portion of the wing surface. As the wing profile (chord wise) changes to converge towards the trailing edge, the airflow attempts to decelerate from supersonic to subsonic speeds. However, this transition is only possible through a chordwise oblique shock wave. Now, this oblique shocks profile will look like the blue line in your diagram above, but will be sitting at some near-mid chord position.
If the flying wing body is increases speed towards sonic velocity of the air, then this chord wise transition shock wave will become stronger and move just in front of the body leading edge. Finally, once the body is moving at supersonic speeds, the transition shock will become a bow shock in front of the leading edge. In addition to this, there will also be two termination shock waves sitting at the leading edge, one for the upper surface airflow and the other for the lower.
The angle of the shock wave profile will be greatest at low supersonic speeds, but beyond the bow stick sitting in front of the leading edge will narrow in angle and so will its spanwise angle. At hypersonic speeds, the distance between the front shock and leading edge will reduce, leading to immense heat energy being impinged onto the body.
Hope this helps


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Do aerobatics pilots do more to account for turbulence in their flying than passenger plane pilots?
Non-aviator here. I was trying to make a comparison when talking to a friend earlier about how when you're trying to do something complicated and delicate, you have to do more to account for environmental factors that are out of your control.
I made the analogy of a passenger plane pilot that can just kind of "push through" most turbulence as opposed to an aerobatics pilot who has to "play the wind" more, accounting for the turbulence instead of just barreling through. It occurred to me, though, that I don't know whether this is actually true. Is my aviation analogy incorrect, and if so, is there a way to make it better?

Best way to make it better is to listen to the guy on the route in front of you and if he says "moderate turbulence", ask for a deviation.

Your analogy doesn't have any real basis to it.  I.e. why couldn't an aerobatic pilot just "push through" turbulence and a passenger plane pilot "play the wind"?  They are definitely different, but your comparison doesn't help us understand the root of your confusion...

If you are flying a faster aircraft, your difference between stall Vs and manoeuvring speed Va is correspondingly larger, so you can dare push through stronger turbulence. And if you are flying a larger aircraft, the larger moment of inertia means the pitching and banking is slower, though so is your control response. But the technique is still the same, just different points where you decide the turbulence is too strong and you need to fly around the weather.


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Which speed should we look for recommended brake cooling in case of rejected takeoff?
Which speed should we look for recommended brake cooling in case of rejected takeoff? (Airspeed or groundspeed)

If the wheels are rolling on the ground, why would airspeed matter?

This is likely to be type dependent. Consult the manufacturer's documentation.

@Frog Because the brake pads are not ground-cooled.

@Sneftel I hear you but E=1/2mv^2 relative to groundspeed.  Air cooling may be just as effective with airspeed perpendicular to direction of travel.  Reductio ad absurdum is that if you choose to take off with a strong tailwind, are you to expect near-zero brake heating?

@Frog yep, fair point.


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Why is the CG assumed to be between the CP and the tail?
While discussing stability (Longitudinal, Lateral, Directional), invariably, all texts assume CG to be between CP and tail.
Ex: Look at the specific image in wikipedia https://en.wikipedia.org/wiki/Longitudinal_static_stability

How ever while discussing Drag, Stall etc, standard text assumes, CG is front of CP as below

Why is this difference?  Which is correct?
Using the first image, I can explain to myself how a gust from down increases lift on wings and gives a clockwise moment which the tail wind moment opposes with anticlockwise - rendering stability.    But I am unable to apply the same argument with second image with CG forward. (Because a gust from down will increase lifting effects of wing and tail as far as I can tell)   Feels like some one is conveniently imagining up to suit their arguments :)
Can anyone explain how longitudinal stability applies when CG is forward and a gust from below happens? Or why 2 different perspectives are used for explaining?

That first drawing is confusing. The checkered circle seems to indicate the CG since the W vector is shown straight down from it. Yet the torque arrow is shown centered around the quarter chord when it should be centered around the CG. In level flight CP is directly above CG. Otherwise there would be a pitching moment. Having CG aft of the aerodynamic center is unusual and unstable. It can be done but it’s usually avoided.

It's important to distinguish between longitudinal stability forces and trim forces, and I find that the literature in general does a very poor job of explaining it for the quasi-layperson.  I use this concept to make it easier to picture what is going on in your mind.
Longitudinal stability is simply the airplane weathervaning about the C of G into the airstream, like the weathervane on a barn turned so that its axis is horizontal.  In other words, an airplane's static pitch stability is simply a horizontal weathervaning effect.
For the airplane to be longitudinally stable, it simply has to be good at this weathervaning action.  This only requires that the overall aerodynamic center of the combined horizontal footprint of wings, nacelles, fuselage, and tail surfaces, the Neutral Point, be aft of the "weathervane hinge", the C of G.
Within that system, you need to have a means to control the actual angle to the relative wind that with weathervane will seek in doing its weathervane thing. This allows you to use a large lifting surface to hold the whole thing up, that you can force to operate at an angle to the airstream to make the lifting force, and keep it at that angle as the relative wind direction changes.  If you don't, you just end up with a lawn dart, which has tremendous longitudinal stability (its Neutral Point is WAY aft of the C of G) but can only go in ballistic arcs.
This is done by creating an opposing force balance within the weathervane; a force seeking rotation about the pivot axis on one direction, opposed by a force trying to rotate the weathervane in the opposite direction.
If you take a weathervane off a barn, which normally wants to point straight into the airflow (airflow parallel to its longitudinal axis), and bend the rearmost part so it deflects air locally to one side, and stick it back on the barn, it will now point into wind with an offset; there is a force pushing the back end out, and as that happens an opposing force builds up resisting that movement until they are in balance and the weathervane will now seek to point into wind at whatever offset angle that the balance point is reached.
You can call that "trimming the weathervane", making it operate at angles off the wind that you can control.  The horizontal tail does the same thing with the airplane in the vertical plane. The offset angle that the airplane will seek is its Angle of Attack.
So trim forces cause a force balance between opposing pitching moments that makes the plane always want to fly with some offset to the relative wind, an Angle of Attack, and static stability means that when the relative wind changes for whatever reason, the airplane will rotate like a weathervane to maintain the Angle of Attack set by the trimming force balance.
I'm not quite sure what the first diagram is depicting (I think its depicting the sum of forces in creating the "weathervane effect"), but the second colour diagram is clearly illustrating the trimming force balance. Tail downforce opposes net nose-down pitching forces that includes the pitching moment of the wing's airfoil, the C of G to Center of Lift leverage, and any other pitch motion inducing forces like high or low offset thrust lines, or things hanging down.  Tail downforce in opposition to those forces controls the offset angle at which the airplane will seek to weathervane itself into the airstream about the C of G.
That "angle seeking" tendency itself is static stability, and is a function of the net aerodynamic center of the airplane-footprint-as-a-weathervane, its Neutral Point, relative to its "hinge axis", its Center of Gravity.
So for good static stability, C of G needs to be forward of the Neutral Point, the Static Margin.  For control of Angle of Attack, G of G is normally placed ahead of the Center of Lift so that an adequate Nose Down pitching moment is generated, that can be opposed by the tail's downforce, to control Angle of Attack.


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How to visualize compass deviation?
What will the magnetic compass which has an Easterly deviation of 10 read when the true heading is North
Will it read 350W or 10E?
I understand that the 10E East variation means that the underlying magnet is swung 10 deg clockwise instead of pointing to true North. So it is now pointing to 10E. But I guess since the readings are done from opposite side of the compass for the end user, the reading shown will be 350W.
Help!
edit : The question is different from what is answered at How do I convert true heading to magnetic heading or the other way around? --  My question is mainly regarding how to visualize the same. - John's answer below is perfect.

I understand the math and what is explained on it - but more confused on how compass mechanism works - esp given compass is read from behind

I assume your question is about Declination/Variation, not Deviation that is addressed with the compass correction card.  It's confusing because you are reading the compass card from behind so to speak and you are rotating around the magnet, which is always aligned with Magnetic North, when you turn.
So forget about the magnet alignment and just consider the compass card indication that is telling your where the nose of the airplane is pointing, and think it through this way:
If you are flying straight toward the Magnetic Pole on the 10 Deg E Variation line, your compass heading will be 360/0 Magnetic, and your True Heading will be 10 Deg, because the nose is pointed 10 Deg to the right of the True North Pole with your compass showing 360/0 Magnetic.
If you are flying straight toward the True North Pole, your true heading is 360/0 True, and your Magnetic heading will reflect the fact that you are pointing 10 degrees to the left of the Magnetic Pole in order to point straight at the North Pole.  This will show as 350 on your compass.
Hence the "East is Least" rule for Easterly Deviation (West of the Agonic Line). To fly a specific True Heading, point so that the Compass indicates 10 Deg LESS (least=less=subtract from True for a Magnetic heading to fly) than your target True Heading; to fly 360 True, Point at 350 Magnetic.
So, per your question, if you are flying 360/0 True, that is, pointing straight at the North Pole, your compass will show 350 Magnetic if you are on the line of 10 Deg Easterly Declination/Variation.

The OP asked about Deviation, and also introduced "Variation" into the question.  These have different meanings in aviators parlance. Your illustration appears to be laying out "Variation" a.k.a. "Declination" and you have labeled it "Deviation", which is usually used to describe magnetic disturbances peculiar the the aircraft.

Good point.  I meant declination and just wrote deviation without thinking about the terminology.  However, his question clearly is about Declination/Variation or he/she wouldn't be asking about true north vs mag north.  If it was about Deviation, it the question would just be about use of the correction card.  Also note it says "variation" farther down. Thanks for bringing it up. I fixed my post.

Thanks for the explanation John.  This is very helpful!


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Why does the X-3 have two oil pressure gauges per engine?
The X-3, as currently in the National Museum of the US Air Force, has two oil pressure gauges per engine, and one fuel pressure gauge per engine. One of the per-engine oil pressure gauges is labeled as being for the turbojet, the other for the afterburner. (But why would an afterburner need lubrication?)
The preliminary flight operating manual says that the engine instrumentation would be the other way around: one fuel pressure gauge for the turbojet part and one for the afterburner part of each engine, and just one oil pressure gauge per engine. What could it be that the two separate oil pressure gauges per engine measure?

In the event no one comes up with an official source, the best guess is almost certainly that the afterburner nozzle actuators were pressurized by engine oil and you needed to know that the pressure was adequate to avoid problems when selecting reheat with the throttle if the nozzle failed to reshape as required.

I'd allow for the possibility that the middle gauges should be fuel pressure gauges, and that the have oil pressure gauges installed where they shouldn't be. Not saying that this is the case, but it seems possible.

@RalphJ, you nailed it. Good instinct :)

After some research, Ralph's comment is almost certainly the answer here: during the restoration they installed incorrect (though period-accurate) oil pressure gauges, rather than fuel pressure gauges. I found a copy of the X-3 Stilleto flight handbook on Google Books, and the relevant pages are included in the preview. Here's a snap of the panel image, Fig 1-12, on page 10:

It's a bit low resolution, but you can tell (by comparing the gauges, at least) that of those six gauges, the top four are the same (reading "FUEL PRESS"), and the bottom two are the odd ones out ("OIL PRESS") -- the opposite of the photo of the panel currently on display.
Item 1-26 has a single entry for oil pressure:

Items 1-40 and 1-41 describe the fuel pressure gauges:


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Did military aircraft in the sixties commonly lack VHF radios?
Investigating the XB-70 cockpit equipment, I notice that it only had (two) UHF radios, of the AN/ARC-50 model, for the 225–399.95 MHz band. No VHF radio that could be used to talk to civilian ATC or civilian aircraft. Is that correct? Was that normal in those days?

Based on https://www.aviationtoday.com/2005/02/01/militarycivil-interoperability/ , it does seem that interoperability with civilian radios was not a priority. Presumably not at all for an aircraft of which only two prototypes were built that were used as experimental aircraft and had a rather short time in service, and never left the US.

A related question is when did the military start using UHF radios? Were WWII aircraft all generally equipped with only UHF?

I can answer for the USAF. In the early 70s, The T-37 and the T-38, (Undergraduate Pilot Training aircraft), did not have VHF, only UHF, and the F-4C/D/E had only UHF.

Same for USMC and USN in the 80s, 90s, and 00s.

The RAF Dominie, from the sixties, had a single-channel VHF radio, a single-channel UHF radio, a single-channel HF, and some listen-only capability on the nav radios. And they could, of course, observe signal lamps.
Actually, that's slightly unfair. The UHF set could monitor Guard and also the selected channel at the same time.

The Dominie was a version of the Hawker Siddeley (later BAe) 125 biz-jet, made for the RAF as an advanced navigation trainer.

Yep, and the equipment fitted to it was completely different to the civilian HS.125

Most of the military aircraft that I flew (
Jet Provost, Hunter, Phantom, Vulcan etc)in the 70’s only had UHF radios, usually an ARC 52 box.  There was no requirement for VHF, the UK had total UHF military ATC coverage as did anywhere else that we were liable to take the aircraft.  You have to bear in mind that airways flying would only have been done by air transport type aircraft which would have had  VHF fitted for that purpose.

Typically, yes that’s true, most military are craft only had UHF radios aboared.  This is not a problem for comms with ATC in the NAS, since they operate on UHF frequencies as well.  It would make it difficult to communicate by radio with civilian airplanes.


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Early turbojet RPM and thrust indicator
The flight operations handbook for the Douglas X-3 Stiletto (an experimental aircraft from the early 1950s) says that the instrument panel has instruments that indicate the thrust of the two Westinghouse J34 turbojet engines. The actual aircraft in the USAF museum has instruments as in the picture. Despite being labeled "RPM", is it so that the outer scale actually shows the estimated thrust of the engine? Presumably calculated using some simple fixed mechanical linkage directly from the RPM.
The J34 model used is indeed said to have produced around 3000 lbf thrust, so that matches. The smaller inner scale shows the RPM, which for this engine goes up to 12,500 rpm, so that also matches.
Aircraft with more modern turbojets or turbofans rarely or never have instruments that directly would show the thrust, right? Is that because such an instrument is highly unreliable anyway and the actual relationship between RPM and thrust depends on factors that a simple electromechanical instrument can't know? (A glass cockpit could calculate it more accurately and display it, but I assume there is little interest in that information?)

Cropped from nationalmuseum.af.mil

It doesn't seem logical that you would label something RPM that is actually intended to represent something else.  Could it be for 2 different spools, and if so could there be that much difference between low and high pressure sections?

@MichaelHall This is an early single-spool turbojet.

@ymb1So that for instance when the rotational speed is 9000 rpm, the outer needle would point to 1, and the inner needle would point a quarter of the way between 8 and 12? Weird. Possible, sure.


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Were the original gyroscopic artificial horizons "upside-down"?
Looking at an article in Flight magazine from March 1945 titled "Sperry Attitude Gyro" that you can find online, the picture of the instrument is interesting:
It seems that the orientation of the sphere was upside-down compared to current convention. The dark "earth" was the top part of the sphere and the light "sky" the bottom. Back in those days, the sphere apparently contained the actual rotating gyroscope, so for instance the last illustration in the picture, 30 degree climb, is how it indeed would show up when the sphere is held in constant orientation by the gyroscope and the aircraft is in 30 degrees climb.
At some point this was mechanism and convention apparently changed, because nowadays there is just a partial sphere surface that obviously does not contain the gyro, and the coupling to the gyro is such that it works the other way in the vertical direction. I guess this indeed more natural, with the "sky" being in the upper part of the partial sphere surface. Am I right?

Attitude indicators (AIs) in Soviet Bloc aircraft operated opposite to the way western AIs worked, and cause numerous accidents in pilots converting to western designed aircraft after the fall of the Soviet Union. In these AIs, the background remained fixed, and the little aircraft moved around. In a dive with a left bank, the little aircraft symbol would move lower, over the dark (ground) background, and roll to the left.

Yes top and bottom were reversed, and there were labels to indicate climb and dive directions. Sperry advertisement circa 1959:

Source
However such indicators were competing with another design from RC Allen, a black background with a white horizontal line:

Source
The white line moved like the AI we know today. So both indicators were showing pitch in an opposed way:

Sperry F3 vs conventional attitude indicator, source
These confusing conventions may have been a contribution in the Beechcraft Bonanza accident in which Buddy Holly died in 1959: See this video.
The Sperry indicator from this video:

The pitch display of this instrument is the reverse of the instrument [the pilot] was accustomed to; therefore, he could have become confused and thought that he was making a climbing turn when, in reality, he was making a descending turn.

Source

Initially I had a hard time believing that the illustrations attached to the question could be accurate. I've added a comment to this related answer https://aviation.stackexchange.com/a/73772/34686  that may help folks understand why the instrument gives "backwards" indications in pitch but not roll. I still feel that this sort of instrument would be incredibly confusing and unintuitive to use.  So, for roll, you are supposed to image that the line on the ball represents the horizon and the fixed bar represents the aircraft, but in pitch you are supposed to image the opposite?  Yikes!

("image" =imagine)


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Strange airspeed indicator in the Douglas X-3 Stiletto
The cockpit of the Douglas X-3 Stiletto has a rather weird looking airspeed indicator. (The leftmost instrument in the top row.) Two concentric circular scales, the outer one goes from 0 to 20, the inner from 20 to 44. What do the scales show?
The actual max airspeed of the single X-3 aircraft built was about 615 knots or Mach 1.0. The initially planned max airspeed was about Mach 2 but that was unobtainable as the intended engines did not become available and lower thrust ones had to be used instead.
Could it simply be that the scales are KIAS * 10, and as the aircraft starts moving, first the needle for the outer scale goes up to 20 and then the inner needle starts moving from 20 up to 44, i.e. 440 knots? (But that is much less than the actual max airspeed.)

Source: nationalmuseum.af.mil

If the 615 is TAS at 20,000ft the IAS might be in the 450 range.

Ah, thanks @Jim, sorry I did not check that. Then this indeed makes sense.


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Steady-state simulation of pitching motion
Is it possible to perform a simulation of 'steady-state' pitching motion of an aircraft/wing? Is there any method where we can analyse the effects through a steady-state simulation of a pitching motion? I have seen numerous papers/articles where the unsteady simulation has been carried out but none related to steady-state.

By definition, a pitching motion makes the process unsteady. Strictly speaking, even straight and level flight is unsteady since fuel is consumed, reducing airplane mass in the process. We assume quasi-steady conditions for straight and turning flight and even for climbs and descents, but at some point this cannot be justified any more.

Okay! Thank you very much for the information.

Do you mean a constant rate of change of pitch, say, +3 degrees per second, maintained for more than just a few seconds?

In fact, constant rate of change of pitch rate (i.e. pitching acceleration is constant).

A "steady" rate of pitch condition is very rare because most aircraft are designed to be stable enough to avoid stalling themselves.  In other words, the more you pitch, the greater the control input you need to continue increasing AOA.
You can reach steady state in roll easily.  Here drag resistance to rolling will stabilize roll rate, provided the aircraft rolls smoothly around its axis.  Here, rudder and elevator inputs may be required.
So, for pitch, you could get a steady state pitch (relative to the horizon) in a loop.  Here, throttle inputs are definitely required (to get a perfect circle).
But for increasing AOA at a steady rate, this will lead to a stall.
There is also the analytical consideration of acceleration before reaching steady state.  Taking the easiest, the roll, roll rate goes from 0 degrees/second when ailerons are applied to steady degrees/second when drag force = rolling force (differences in AOA of rolling wings helps too).
Many early "video arcade" games do not simulate the acceleration (the deer is instantaneously in full gallop). To do an accurate simulation, acceleration to steady state must also be considered, so now you have a rate of acceleration to steady state.

Thank you very much for such an informative answer! I understood much of what you have mentioned and I could get a feel too of what you are suggesting. However, the last paragraph seem to be little confusing and hence I am yet to comprehend it.

@angryam we start with a steady state of no motion, and end with a "steady state" of constant motion.  Interesting in that "steady" can be considered in terms of distance, velocity, or acceleration.  But to accurately simulate a pitching action, we must consider the acceleration to steady state velocity (with applied force).  In air, it is tricky because drag opposes applied force as speed increases.


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Spar-rib-stringer spacing and their thickness in relation to the wingskin thickness
I am solving a wing structure for normal modes using SOL103 in NASTRAN. Since I am beginner, I started with just the wingskin (lower and upper wingskins) but that results were far from what was expected. Only the local modes of the wingskin was observed. I modelled the spars and ribs then the 1st mode of bending was observed, rest all were the local modes of wingskin. Increasing the number of ribs caused the 2nd mode of bending to appear. Introducing stringers stiffened the structure where the 1st twisting mode and bending in x-y plane (my model has chord along x-axis and span along y-axis) were obtained. However still the local bending modes of the wingskin are seen where seen. So what is the appropriate spar-rib-stringer spacing and  their thicknesses and the wingksin thickness so as to obtain the expected normal modes of a wing?
My wing is a sweptback wing with 30deg sweep angle, taper ratio of 0.6, span of 2m, and made of NACA0012 airfoil. I intend to simulate the cruise condition.

It needs to be stronger than "cruise condition".  Combinations of materials good in compression and those good in tension are helpful (starting with wood and cable!).  Fiber glass over foam core should also yield some clues.  Avoid harmonicly reinforcing bending modes.

@RobertDiGiovanni SOL103 is linear eigenmodes so no difference in compression/tension stiffness (let alone strength) can be modelled. The choice of materials is not the main issue here.

Why would you combine 30° sweep with a NACA0012? What speed are you designing for?

It seems like you discovered the inconvenient truth about thin-walled structures. Despite their apparent superiority (all the material is as far away as possible from the bending line) they are extremely flimsy. This is because the ratio of out-of-plane stiffness and mass is very poor: flexural rigidity $D\propto t^3$, mass $m\propto t$. Especially when using materials with a high volumetric stiffness (e.g., steel), the resulting structure becomes so thin that the plates have unpractically low eigenmodes. The same goes for buckling under flight loads.
Instead, the aircraft designer must resort to a poorer volumetric efficiency to get sufficient out-of-plane strength and stiffness. Basically, increase that $t$ without increasing $m$ too much by choosing perhaps less stiff but much lighter materials (welcome composites and perhaps aluminium). However, that still leaves (too) much to be desired, so you add stiffeners, stringers, ribs... (all increasing the effective $t$) until a sufficient performance is obtained. Most efficient is to make a so-called sandwich construction: two stiff layers with a light (e.g., foam) core in-between, which has a stiffness of $D\propto t\cdot h^2$ (with $h$ the foam core thickness and $t$ the skin thickness). See e.g., here.
This is all to say that there is no optimal rib/stringer spacing unless you know how the 'skin' is constructed. Remember, eigenmodes are all about stiffness and mass ($f\propto\dfrac{k}{m}$), so low eigenmodes are indicative of mass not contributing to stiffness of that particular mode.

A very interesting and informative answer. Thanks!

Seems like adequate strength can be had with thicker skin, and the foam core acts to damp unwanted oscillations (but is very light).

@Robert No the main effect here is stiffness (force due to displacement), not damping (force due to velocity). Although of course the damping is a welcome side effect, and the modes are especially easy to damp now we raised their frequency. Of course, this won't show up in a SOL103 analysis which only looks at real eigenvalues.

to obtain the expected normal modes of a wing

One might turn to nature to get a better feel for this issue.  One may build strong and stiff, but it will be heavy.  And even skyscrapers have harmonic modes.
The natural solution is a combination of strength and shock absorbing ability.  The bone and tendon arrangement, seen in the horse, allows 1000 lbs of weight to be dropped on one foot over and over again without breaking the bone.
The tendon aborbs the impact over a longer period of time, reducing strength requirement.  What is worrisome about adding internal structure (such as ribs) and thinner skin is that it will create a levered bending point (at the rib).  So one part is too stiff, the other not stiff enough (thin skin), so to reduce fatigue at that point, more ribs must be added.
It seems a flexible, shock absorbing approach serves to not only reduce peak strength requirement (from a gust or manuvering),saving weight, but also will reduce or eliminate unwanted harmonics.
So now we may look at another design, avian wing bone.  Even without the attached feathers, muscles, ligaments, and tendons, we see a round, thick walled structure with a hollow interior.  Strong, flexible, and light.


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What are the different types of take-off minima at the bottom of a Jeppesen airport diagram?

Anyone that could explain the bottom plate, what’s the difference between all those minima?

The used acronyms are defined in the Jeppesen charts legend:

CL Centerline Lights
RCLM Runway Center Line Markings
RL Runway (edge) Lights
RVR Runway Visual Range

The first column tells us that a takeoff is authorized with RVR in touchdown zone (TDZ), midfield (MID) and rollout (RO) of at least 150m, if the RL and CL are operational (except during snow, where 400m is required, and for runway 18, where 250m is required).
The second column tells us that takeoff is authorized, if the RVR is at least 200m with RL and CL operational. In this case not all RVR values need to be available (e.g. one of them could be inop).
The third column tells us that takeoff is authorized, if the RVR is at least 300m and RL and RCLM are available (CL is not required, except at night).
The fourth column tells us that takeoff is authorized, if the visibility is at least 400m and either RL or RCLM (day) / RL or CL (night) are available.
The fifth column allows taking off with adequate visual reference during the day with at least 500m visibility. The "adequate vis ref" is defined in the charts legend as:

ADEQUATE VIS REF (Adequate Visual Reference) — Runway markings or runway lighting that provides the pilot with adequate visual reference to continuously identify the take-off surface and maintain directional control throughout the take-off run.


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How can a large passenger airline in midair hover motionless for 10 minutes above my house? This happened in 12/2020
A few weeks ago, sometime in December 2020, I walked outside to get my mail around 1300 and saw a large passenger plane hovering roughly 5000ft above my house. I watched it for 5 minutes when a UPS truck came to deliver a package to a neighbor.  I told them to look up to see what I was seeing.  We watched in amazement. Five minutes later, I noticed the plane start to inch forward. I watched it for another 15 minutes and it couldn't have been more than 5-10 miles away from me.  All this happened on a clear blue sky day, with just a slight breeze at ground level. I called the airport to see what they had to say and only got a recording to leave a message, so I did.  I never heard back from the airport tower, but I could not get the incident out of my mind. This occurred over Pasco, WA USA, and I would swear on a Bible that three of us saw the same phenomena. I know how plans can fly, and there had to have been one heck of a wind event for this to happen. Hopefully, someone can give me a reasonable explanation for what I saw. Marja Henderson, retired RN (1/17/2021)

That isn't possible, so whatever you saw was obviously not a large passenger jet.

How did you arrive at your estimates of 5000' and 5 -10 miles? Only possibility I can think of is a scale model . . . . what about engine sound?

According to FlightRadar24, SkyWest Flight 3830 took off from the Pesco Tri-Cities Airport at about 13:06 local time, flew directly over the city, then climbed away to the southeast (note: FlightRadar24 uses UTC time, so 13:00 local would be 21:00 UTC). You would have seen it directly tail-on as it climbed to its cruising altitude.
It's well known among pilots that, if you're on a collision course with another aircraft, then the aircraft will appear stationary in your field of view. So, if you see an aircraft that appears to not be moving, you better pay very close attention to it until you determine if you need to take evasive action or not.
I think what you saw was SkyWest 3830, but directly tail-on, so you wouldn't have seen any movement across your field of view. The fact that it was so far away meant you couldn't tell that it actually was moving directly away from you.

That's not compatible with alleged observations: "Five minutes later, I noticed the plane start to inch forward. I watched it for another 15 minutes and it couldn't have been more than 5-10 miles away from me". At the slowest acceptable airspeed, the aircraft would have disappeared from view.

@mins Not true. Even at cruise speed, a passenger jet could have remained in view for half an hour or more, and SK@3830 wasn't flying anywhere near cruise speed at that time. Five minutes after takeoff, it was only at about 20,000 feet and still climbing. After that point, the rate of climb slows (as would be expected from the thinning atmosphere) but forward speed doesn't, which results in the start of the apparent forward motion (because the plane is no longer flying directly away from the observer).

"I watched it for another 15 minutes and it couldn't have been more than 5-10 miles away from me". At, say, a slow 200 kt GS, in 15 minutes the aircraft travels 90 km, and is out of the sight, first by the effect of nebulosity/diffraction, and second because its angular size, smaller than the one of ISS at 400 km, is now below human eye resolution.

There is an air force base near Spokane which is home to large jet planes, at which pilots train to fly them. It is possible that the pilots of that plane were practicing slow flight, where they set up the plane to stay in the air at the minimum possible (safe) speed. A big jet like a B-52 can be slowed to around 140 knots and remain just barely airborne, and in a 100MPH headwind the plane will be moving across the ground at 40 knots. If viewed from a mile away, it will look as if the plane is hardly moving at all.
In addition, if the pilots were practicing stalls, where the plane is going so slow that it is teetering right on the verge of falling out of the sky, they can "hang" the plane in the air by piling on full power and tilting the nose up so the engine thrust is helping maintain the plane's altitude at that (dangerously!) slow speed.
I invite the B-52 experts here to weigh in with their experiences regarding this.


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What causes a taildragger to lift its tail during the takeoff roll?
The rear of a taildragger is closer to the ground than the front. As the aircraft moves forward (I guess at its stall speed), its tail rises up and the aircraft moves with the fuselage level before takeoff (As seen here). But why? Does the pilot change the elevator?

Yes, when the plane exceeds a certain speed (known to the pilot!), the pilot then pushes the stick slightly forward and this lifts the tail off the ground. This reduces rolling drag and sets up the angle of attack of the wing for takeoff. For very powerful airplanes, the tail will sometimes come up on its own and surprise the pilot during the takeoff roll.
In some planes, the resulting angle of attack is enough that the plane will fly itself off the ground and begin climbing when it has picked up enough speed in the roll. In other designs, the pilot will wait for the right speed and then gently pull back on the stick to increase the AoA a bit and the plane then takes off.

"For very powerful airplanes, the tail will sometimes come up on its own" -- so what causes it in this case?

@nanoman, the prop wash acting on the elevator surfaces, with the ground close by, can lift the tail off the ground. This happened in the P-51 Mustang if the pilot did not hold the stick back on his roll until he was properly lined up to complete the roll on the main gear, at which point he would then push the stick forward a but to lift the tail on purpose when he was ready for it.

@nanoman, are you familiar with weathervanes?  The tail has a lot of surface area and it wants to align with the relative wind.

Think of it as a wheel barrow, with the center of gravity behind the forward set front wheels.
The weight x distance of the CG behind the front wheels (when the aircraft is at rest) creates a torque force holding the tail down.  As the plane begins it take-off roll, the tail and wings begin to counter-act the weight torque with lift, raising the tail off the ground.  In the video, the pilot "helped" it a bit with some down elevator (increases tail lift coefficient), easing off as speed increased.
It is important to note, with proper technique and design, the plane should not be at an angle of attack to fly off at the lowest possible airspeed during the take-off run.
Just like a tricycle gear, angle of attack should be held lower until the aircraft is moving fast enough to lift off safely even in the event of a change in wind.  The concept of rotation to fly-off is the same, around 1.2 Vstall for a recreational aircraft.

In some tailwheel aircraft, at some trim settings, if the pilot exerts no forward or aft pressure on the control stick or yoke, the tailwheel will rise off the ground as the aircraft picks up speed during the take-off run, and if the pilot continues to exert no forward or aft pressure on the control stick or yoke, the aircraft will rise into the air at a reasonable flying speed at an angle-of-attack that is safe for normal conditions.  After the tail is up, if the pilot wishes to attain a higher than normal takeoff speed, he may exert forward pressure on the stick or yoke to keep the main wheels on the ground for an extended time, so that the plane eventually lifts off at a rather low angle-of-attack.  Conversely, by exerting aft pressure on the control stick or yoke, the pilot can make the plane lift off of the ground from the "three-point" attitude, so that all three wheels leave the ground at the same time.  This maneuver should be approached with caution due to the high angle-of-attack involved, but it is safe in some aircraft in smooth conditions.
As to why the tailwheel rises off the ground if the pilot exerts no forward or aft pressure on the control stick or yoke-- it is basically because the plane is trimmed to fly at a much lower angle-of-attack than the angle-of-attack that exists when the aircraft is travelling horizontally along the ground in the "three-point" attitude, with all three wheels on the ground.  As the weight on the wheels decreases during the takeoff run, the aerodynamic forces become more and more important in comparison to the forces exerted by the wheels, and the plane tends to adopt an angle-of-attack close to the angle-of-attack that it is trimmed for.


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Can a plane with a thrust-to-weight ratio less than one fly vertically and perform loops?
In the game Battlefield 1 you can fly a Sopwith Camel vertically and perform different kinds of aerobatic maneuvers like loops. According to its specifications, its thrust-to-weight ratio should be about 0.28 and hence it should not be able to fly vertically. Can it?

If you go into a dive you can pick up speed and then loop. However, sustained vertical flight should not be possible.

With thrust/weight ratio above one, you can use the thrust of the engine to accelerate upward (at least at some point in the flight envelope).  As long as the engines can deliver full thrust, you can maintain that position.
Below one, vertical flight is still possible, but you will be losing speed in the maneuver.
Gliders are capable of loops without any thrust at all. The only energy in the maneuver is the speed of the glider on entry.

Yes.  Plenty of airplanes with thrust to weight ratios less than 1 can perform vertical maneuvers though time in high pitch or vertical lines is limited as airspeed drops rapidly.  Often such airplanes have to be placed in a shallow dive to accelerate them to an appropriate entry airspeed for the maneuver.  This is sometimes true for aircraft which have a thrust weight ratio greater than one as well, as, while a thrust to weight ratio > 1 provides an expanded maneuvering capability in the vertical, the aircraft is still slowing down as the combined force of the weight of the aircraft plus drag on the airframe are generally greater than engine thrust.


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(UK) Why did Colin Furze's Hoverbike not require a licence?
Link:

Why did he not need a licence to fly this?

Who says he didn't?

Maybe because they consider it a ground effect vehicle and nobody in their right mind would ever take it out of ground effect.

@WPNSGuy He hasn't mentioned any sort of licence in any of his videos at all

@JohnK Colin Furze lives in the UK. According to this article, the UK considers ground-effect vehicles to be airplanes.

Colin's been in legal trouble with a number of his builds -- what makes you think he didn't get some hassle from this one?

What I recall from the video (I only watched it once, when it first dropped), one of the challenges of this build was making the machine light enough to meet UK/EU requirements for a no-license ultralight -- the rectangle-tube aluminum frame, minimal seat and handlebars all went toward this goal.
Bottom line is, he's below the UK weight limit, below the maximum speed permitted, and has less than the maximum fuel capacity allowed, so it's legally a no-license ultralight under UK/EU regulations.
Also, this build was sponsored by a major car manufacturer (Ford Motor Company, as I recall), who might well have helped with legal research as well as funding.

I would add that this contraption,  with all due respect, is pretty much certainly unable to operate outside ground effect. This may also free it from regulation. It is essentially a hovercraft of less than 1000kg mass, and as such not subject to registration in the UK.

FWIW, my recollection from the video is that this would also qualify as an ultralight in the US under Part 103 -- but might need to be registered as a boat like other hovercraft usually are.

It could if the HP was high enough.  You just wouldn't want to get it high enough that the highly probable 2 stroke engine failure  turns you into a falling object.


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Could "Lady Be Good" Incident Prevented by the Technology of its Time?
For those who don't know, Wiki page of the incident sums up. I don't have much knowledge about communication systems of WWII, but I know how radio communication work. This page has

The aircraft flew on a 150 degree course toward Benina Airfield. The
craft radioed for a directional reading from the HF/DF station at
Benina and received a reading of 330 degrees from Benina. The actions
of the pilot in flying 440 miles [710 km] into the desert, however,
indicate the navigator probably took a reciprocal reading off the back
of the radio directional loop antenna from a position beyond and south
of Benina but 'on course'. The pilot flew into the desert, thinking he
was still over the Mediterranean and on his way to Benina.

After reading, I thought that there could be a way to save those men by using a right method of radio communication.

The issue here is that the HF/DF (High Frequency Direction Finder) at Benina had only a single loop antenna, with a symmetrical gain. This means that a signal from an aircraft at a bearing of 150 degrees (where the aircraft actually was) and a bearing of 330 degrees (where the aircraft was told it was by the radio operator at Benina) look exactly the same.
The on-board ADF (Automatic Direction Finder), which would have kept this from being a problem in the first place, was reportedly inoperative.
Thus the pilot believed that he was on course to Benina, when in fact he was on course directly away from Benina.
Had the pilot been aware of the possibility that he had already overflown the base, there are ways to disambiguate the result. For instance, he could have changed course to the right and asked for a new reading. This would show as a course change to the right if the navigator were reading the correct bearing and as a course change to the left if he was reading the reciprocal heading.
In the situation where the pilot stays on course, there is no way to fix this using technology. At some point, the pilot could have realized he had been flying too long to possibly still be on course to Benina. Lacking this sort of realization, the flight was doomed.

The Wikipedia page mentions "flares fired to attract their attention" which suggests that somebody at the base knew that they'd overflown it; whatever breakdown prevented that knowledge from reaching the crew probably contributed to the accident. Also, the Wikipedia page states that the ADF in the aircraft was inoperative; that is "technology of its time" that absolutely would have prevented the accident. Sadly, equipment of any era can fail, sometimes with tragic results.

@RalphJ I'm not so sure the flares mean they knew it had flown over. The fact that the search took place in the Mediterranean instead of the desert indicates nobody did in fact know that... Perhaps they were just firing flares periodically?

Fair point; without knowing more about why the flares were fired, it's hard to know if a really key piece of the puzzle was known but ignored (i.e. they'd overflown, but radio operators & SAR bosses didn't believe the report), or if that piece simply wasn't there at all (nobody knew they'd overflown & the flares where just periodic).


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Do vortex generators increase lift even at low AoAs?
Does wing produce higher lift with VGs even at low AoAs(let say 10°) compare with wing without VGs?

First of all, for a straight wing with a cambered airfoil, 10° is not a low angle of attack anymore.
At all flight speeds vortex generators intensify the mixing of high-energy outer air with the slowed-down air of the boundary layer, thickening the boundary layer in the process while keeping flow speed at the surface up. This helps to delay flow separation (which is defined as airflow at the surface coming to a standstill from the deceleration caused by an adverse pressure gradient).
In poorly designed wings those vortex generators also help with control surface linearity and effectivity. A good example for the clever application of vortex generators is the horizontal tail of the Short Belfast, where the vortex generators hide in the elevator hinge gap at small deflections but will protrude into the airflow once the elevator is moved to larger negative deflections, helping to increase the linear range of the elevator (independent of angle of attack).

Horizontal tail of a Short Belfast (own work)
Low Speed
Friction is proportional to the speed gradient at the surface. Keeping this surface speed up will obviously increase friction drag. However, it will also help to shift flow separation to higher angles of attack, so the linear part of the lift curve slope will reach to higher angles of attack. This reduces the minimum flight speed by allowing attached flow at slightly higher angles of attack compared to a wing without vortex generators. But that will only make a difference in lift close to stall and not at lower angles of attack. Drag, however, will be higher over the whole angle of attack range.
High Speed
Vortex generators on the outer wing of a Boeing 737, for example, help to fix the location of the shock by preventing flow separation past the shock. Without the vortex generators, the shock location and intensity would fluctuate and with it lift, leading to an effect called "buzz", resulting from the high-frequency lift variation.

Vortex generators on the outer wing of a Boeing 737-800 (picture source)
This happens already at low angles of attack (typically 3° to 5°) and does indeed increase lift in the low-lift part of the cycle of lift variation. This effect is small, but by eliminating the cyclic loading makes a big difference in flight comfort and the wing's fatigue life.

Can I increase lift at straight wing with big vortex/strakes/generators that produce big strong vortexes, so all upper surface will be coverd with big swirling flow/"tornados"? Big mean,much bigger than boundary layer height, just like fighter jet has vortex from strakes.. I am talking for AoAs where flow is still attached to surface, so basically I am comparing laminar flow vs big vortex flow  over upper surface of wing.I know this will produce huge drag, but I am  now just interested for lift.

@user207141 This would be a hugely inefficient way of getting extra lift. Better to add a flap, ideally with a gap and some chord increase on deflection. This can be folded away for cruise while the VGs will be exposed all the time.

high lift with tons of drag. Joseph Katz work a lot on that. pages 47-50 https://viterbik12.usc.edu/wp-content/uploads/2019/06/Luhar_Aerodynamics.pdf

@user207141 the vortex generators on race car underbodies are more like diffusors than simple aviation-type vortex generators.

Why like diffusors? Isnt they more like strakes at fighter jets which deliberately produce strong vortex flow?     https://imgr1.auto-motor-und-sport.de/Red-Bull-F1-Technik-Unterboden-GP-Monaco-2022-169Gallery-a7c10cbb-1905602.jpg

@user207141 they channel the flow away from the bottom, like a hovercraft in reverse. A diffusor is maybe not the best comparison since those slow air down and increase pressure, while the structure on racing car bottoms will increase suction. Vortex formation is less significant.

These strakes/fences are passive device, even between inner fences and bodywork air is accelerate, I dont see a way how this concept will increase overall mass flow rate in these venturi chanells?  To me its seems that is better without them, using just  clean fresh freestream air, but maybe must be here to manage tire wake,not let dirty air under the floor..

I believe they do if the vortex generators are big enough. The most extreme case of this is the vortex lift on a delta wing at high angles of attack, like the concorde. this article https://ntrs.nasa.gov/api/citations/19930087240/downloads/19930087240.pdf provides some experimental data


text:
What is viscous induced drag?
In this video he talk at 17:04 about viscous inudced drag, induced drag comes exclusively  from inviscid theory,by the definition this drag exist only in inviscid world, so what is viscous induced drag?

There's no such a thing as a "viscous" induced drag and that terminology should be avoided.
Drag coefficient for the entire airplane can be approximated with the well known expression:
$C_d=C_{d_0}+kC²_l$.
The first term is called "parasite drag" while the second term is called "induced drag".
The main source of induced drag is the "drag due to lift" due to pressure forces.
The main source of parasite drag is the "skin friction" due to viscosity.
Anyway, pressure is also responsible for a (normally negligible) fraction of the parasite drag. And viscosity is also responsible for a (normally negligible) fraction of the induced drag. In short:

viscosity
pressure

parasite $C_{d_0}$
skin friction
drag due to stall, wave drag

induced $kC²_l$
profile drag
drag due to lift

This latter should be what in the video is called "viscous induced drag" i.e. the part of the drag which is proportional to the lift and caused by the viscosity.
This drag is simply what is normally seen in a plot of the $C_d$ vs. $C_l$ for a generic 2D airfoil (plot source):

Unfortunately enough this drag is more or less proportional to the square of $C_l$ and is therefore grouped together with the "drag due to lift" to form the "induced drag" term $kC²_l$.
I said unfortunately because this "viscous induced drag" is actually not "induced" (by the lift) and has absolutely nothing to do with the production of lift: it is just a drag due to the growth of the boundary layer with the speed and the angle of attack. It is more correctly termed "profile drag" because the growth of the boundary layer virtually modifies the profile of the airfoil seen by the airflow. I have no idea why a new fancy terminology has been used in that video.

In a nutshell, when he talks about induced drag in the viscosity-free inviscid world, he's talking about the energy consumed in overcoming only the inertia of the air molecules to move them down from point A to point B in generating lift.
It's treating each molecule as an independent object in a vacuum, completely unaffected by adjacent molecules. You could say the inviscid world is only concerned with the energy required to move a single molecule from A to B in a vacuum, then applying that energy value to all the molecules, as if the molecules were coated with a super-lubricant that eliminated friction between them.
But since the molecules are actually packed together as a gas of a certain density, the molecules have to get past each other, inhibited by their own intermolecular friction, or resistance to shearing forces. To make lift in the real viscous world, energy is consumed overcoming the shearing forces in addition to energy consumed overcoming inertia.
So, what he's doing is calculating the energy to make the air move down solely to overcome inertial forces, without accounting for shearing or viscosity effects, the inviscid induced drag, then adding the energy required to overcome viscosity effects, the viscous induced drag, arriving at the total induced drag.

Remember how in 2D airfoil data -- like in Theory of Wing Sections -- the 2D drag looks like a parabola?  Here he is bookkeeping that drag in with the induced drag.
It is a lift dependent drag -- that takes a parabolic form -- so this is reasonable to do.  Furthermore, if you were measuring drag in a wind tunnel, it would be very difficult to separate that drag out from the true induced drag.
However, it is not a true induced drag (in terms of its physical origin).


text:
Help with calculate forces from pressure distribution

V=50m/s
deltaPg= -981 Pa
deltaPd=490,5 Pa
c=1m

r.com/PCshZ.png

Above is calculate Fz with integrals and with average pressure.
Why F3 is calculate from 0 to C/2 if pressure distribution is drawn at bottom surface from X to C/2?
What is pressure distribution at bottom surface from 0 to X?
If I know pressure distribution around airfoil, then shape of airfoil and airfiol AoA(inclination on diagram) is not relevant when calculate forces?
Why pressure is not drawn perpendicular to airfoil surface?
Can you explain how calculate lift,drag,total force and read this pressure distribution diagrams?

I chink you’re reading it wrong.  The pressure plot is based at the zero line not the lower wing surface.

@Jim Yes I read wrong, I look at arrows at surface of airfoil, no arrows at lower surface from 0 to X, so I ask my self what is with pressure here! So airfoil picture in this diagram is completly irrelevant, I must look only at pressure distribution like is based on zero line?What if diagram look like this https://lh3.googleusercontent.com/proxy/arJjevORv-SIAW3uBU0byJhfWRbQabEvK7Ctr16dIYBklCxH_LryHktuGGLpYPxWZUgpmX0ZUPc-7PVGHcDz7ZuM74Q_5TEDRlQbgA

Your question neglects span. Given are pressures (force per area) but you only have an airfoil chord, no spanwise dimension. Only the area produced by chord times span will enable you to calculate a force.

@PeterKämpf Yes  Fz is force per unit of span.

Can you explain how to calculate lift, drag, total force and how to read this pressure distribution diagram?

Not with the given information. All what can be calculated is the lift per unit of span. Pressure is force per area, so in order to arrive at a force, it must be multiplied by an area. We only have chord, no span, so there is no area given. All I can do, therefore, is to calculate the lift force per unit of span. Drag cannot be calculated at all.
Let's take 1 m as this unit: This will give us 1 m² as the wing area. Since the pressure over chord is either constant or linear, no fancy integrals are needed. The plot already distinguishes four sections, each with their own gradient of pressure:

Upper forward: Constant pressure of -981 Pa which equals a suction force of 981 N per m². Per meter of span this is 490.5 N of upward force of the half square meter here.
Upper rear: The pressure grows from -981 Pa to 0 at the trailing edge. Therefore, lift per meter of span is half of what it is in the first section, namely 245.25 N of upward force.
Lower forward: Again a linear increase over chord, now from 0 to 490.5 Pa pressure at mid chord. Per meter of span this is 122.625 N of upward force.
Lower rear: Same thing, only in reverse: Pressure drops from 490.5 Pa pressure at mid chord to 0 at the trailing edge. Again, we have 122.625 N  of upward force per meter of span.

If we sum up all four sections, the result is 981 N of lift per meter of span. Suction on the upper side contributes ¾ of this force and pressure on the lower side adds the fourth quarter. Funny how this coincides with the usual approximation of gravitational acceleration! Each meter of span lifts 100 kg of mass in Earth's gravity field.

Why F3 is calculated from 0 to C/2 if the pressure distribution is drawn at the lower surface from X to C/2?

X is a variable which starts at 0 at the root of the X axis. So F3 runs from X=0 to X=c/2. F3 is the force produced by the forward pressure ramp, nothing more.

What is the pressure distribution at bottom surface from 0 to X?

X is variable. At X=0 the bottom pressure is 0. It grows to 490.4 N/m² at X=c/2 and drops again to zero at the trailing edge when X=c.

If I know the pressure distribution around the airfoil, are the shape of the airfoil and the airfoil AoA (inclination on diagram) not relevant when I calculate the forces?

Yes, the shape and AoA are not relevant (or is this: No, the shape and AoA are not relevant?) German and English answer negated questions in opposite ways when meaning the same. Try to avoid negated questions when you want a clear answer.

Why is the pressure not drawn perpendicularly to the airfoil surface?

This is to simplify the calculation. You can either integrate the cosine of the inclined pressure over the actual length of the airfoil contour, or you can integrate the absolute pressure over the projection of the contour on the X axis. Both will give the same result when you want to find the lift force.

"Not with the given information"  I have question  where this can be explanied: https://aviation.stackexchange.com/questions/90710/how-calculate-lift-drag-using-static-ports-holes-in-wind-tunnel-test

+1 for "Funny how this coincides with the usual approximation of gravitational acceleration! Each meter of span lifts 100 kg of mass in Earth's gravity field." alone

Why F3 is calculate from 0 to C/2 if pressure distribution is drawn at bottom surface from X to C/2?

The pressure distribution is uniform from 0 to c/2, then uniformly different from c/2 to c. It does not start from point $x$, it starts from point $x$=0. $x$ is the horizontal co-ordinate.

What is pressure distribution at bottom surface from 0 to X?

Increasing linearly from zero to $\Delta P_D$

If I know pressure distribution around airfoil, then shape of airfoil and airfiol AoA(inclination on diagram) is not relevant when calculate forces?

The shape of the airfoil at this AoA is what results in the pressure distribution.

Why pressure is not drawn perpendicular to airfoil surface?

The picture seems to be relevant for upwards pressure only.

Can you explain how calculate lift,drag,total force and read this pressure distribution diagrams?

This pressure diagram is relevant to lift only.


text:
What is the mathematical relationship between Specific Excess Power and an aircraft's acceleration at a given point on an E-M diagram?
If an aircraft is on the x-axis of an E-M diagram plot (zero turn rate) where the Specific Excess Energy (Ps) = 800, we understand this to mean that the aircraft has the potential to either instantaneously gain 800 ft/sec of altitude OR instantaneously accelerate.  If an acceleration rate is desired, this could be determined by converting the 800 ft gained (h) as potential energy to kinetic energy using an equation V = SQRT {2 x g x h}.  Substituting the variables, V = SQRT {2 x 32.2 x 800} yields 227 ft/sec gained each second (or acceleration of 227 ft/sec^2)  However, this is an unusually high amount of acceleration for even a high performance jet accelerating in straight, level flight (7 g's).  There is something clearly wrong with my logic/math but I can't figure it out.  Can someone offer the correct derivation?

You forgot that you accelerate not from standstill, but a considerable speed. So that equivalent potential energy change comes on top of the existing kinetic energy and makes the energy change look much smaller (resulting from the square of speed in kinetic energy). The correct calculation is
$$\frac{\delta E_{pot}}{\delta t} = 800 \;\text{ft/sec}\cdot g=(v_1^2-v_0^2)\cdot \frac{1}{2\cdot \delta t}$$
Your acceleration will decrease as speed increases. If I assume a speed of 800 ft/sec, the resulting acceleration is to a speed of
$$v_1 = \sqrt{v_0^2+2\cdot g\cdot h} = 831.55\;\text{ft/sec}$$
within one second which is 31.55 ft/sec² or just shy of 1 g. The initial acceleration is 1 g and decreases as the aircraft picks up speed. In mathematical terms:
$$\ddot x = SEP\cdot\frac{g}{v} = v_z\cdot\frac{g}{v}$$
When climb speed (or specific excess power) equals flight speed, both cancel, leaving gravitational acceleration equal to horizontal acceleration if all climb potential is used for increasing speed.


text:
Can an aircraft flying VFR with a squawk code enter adjacent class C or D airspace after being handed over to the destination tower?
In the US, a VFR aircraft request VFR flight following and is given a squawk code after takeoff.
15nm before arriving at it's destination class D airport, the TRACON facility instructs the pilot to "contact the destination tower".
The pilot keeps his squawk code and change frequency to contact the destination tower. The tower instruct the pilot to enter "straight-in" for the runway in use. However, in order to enter straight-in, the pilot must briefly transit through the class C shelf of an adjacent airport and through the class D of another airport not far from the destination.
Q: Is the pilot cleared to transit through both adjacent C and D airspace since he still squawking the code provided by the local TRACON?
I understand the legal requirements to enter each airspace (AIM 3-2). If the pilot was still in contact with the TRACON facility he would be cleared to enter adjacent C and D airspace since he is squawking a discrete code AND previously established two-way communication. In this case however, the pilot is not anymore talking to the TRACON and switched to the destination tower. I cannot find an official reference that covers this particular scenario.
I came across this question/answer, Does Flight Following automatically include permission to fly through class C or D airspace?, before posting but my scenario is different since the pilot is not talking to the TRACON anymore. It's subtle but makes a big difference.

@Pondlife I came across this question/answer before posting but my scenario is different since the pilot is not talking to the TRACON anymore. It's subtle but makes a big difference.

@randomhead (ctd) -- anyway, the TL;DR version is that I'd be alarmed to learn of any sources that were giving the pilots the idea that communication requirements to enter a Class C "shelf" were different in any way from communication requirements to enter a core area of Class C airspace that actually extended all the way down to the surface.

@randomhead thanks for commenting but "probably" is rarely a good answer in aviation.

@quietflyer the issue has to do with who the "ATC facility providing air traffic services" is (91.130). For the most part: in the outer shelf (and beyond) it's the radar controller, in the surface area (at least up to a certain altitude!) it's the tower controller. Things are even more distinct if the facility is not an up/down. But pilots don't know the exact airspace owned by each controller, hence the "probably." If the pilot was previously in communication with the radar controller when outside the C surface area...

...then they probably wouldn't get in a huge amount of trouble if they were transferred to the D tower controller and got instructions that had them brush the C outer shelf. But brushing the C surface area would be more problematic. Whichever controller is issuing instructions is supposed to coordinate entry into other surface areas, but that doesn't relieve the pilot of their responsibility to follow 91.130 and 91.129.

@LobsterBaz the safe and by-the-book answer is no, unless explicitly told otherwise, the pilot is not permitted to pass through other C/D airspace on their way to the destination D airport. But local ATC procedures and inter-facility letters of agreement may allow for it, possibly.

@quietflyer one more response to "communication requirements to enter a Class C "shelf" were different in any way" — consider a pilot overflying a Class C at 1500 AGL vs 500 AGL. The higher pilot will enter the shelf, which is controlled by the radar position, and should talk to them. The lower pilot will pass below the shelf and will not enter C airspace until they get to the surface area which is controlled by the tower position. (Though in my experience we still want them to talk to the radar controller first, so everyone knows what's up.)

@randomhead --re "the issue has to do with who the "ATC facility providing air traffic services" is (91.130)" -- that's an interesting point, thanks for adding that.  But surely the FAA doesn't construe that once a pilot has legitimately met the requirements to enter a Class C shelf, he or she is then responsible for knowing that he or she needs to speak to a different entity before flying over the part of said airspace that actually extends down to the surface-- do they?

@randomhead -- this may well evolve into a new ASE question -- but in actual practice, does a pilot ever receive instructions to the effect of "communications established, so you may enter the Class C shelf, but establish communications with (other entity) before entering the Class C Surface Area"?  That would be a new one on me if so, though I can see how it could be theoretically possible.

@quietflyer in practicality the answer to both your questions is no... the distinction is essentially irrelevant once communications have been established with the radar controller. It's more important for initial contact, or after contact is terminated, and/or if the tower and radar facilities are distinct (not an up/down).

Trust, but verify.
The pilot must comply with 91.129 and 91.130 unless otherwise instructed by ATC.
ATC of any kind (radar or tower, Class C or Class D) should coordinate with any affected tower if an aircraft they are providing services to will enter another facility's surface area airspace. At least that's how I read the 7110.65 2–1–16, which you will note is in the "General Control" section and not the "Radar" section... though the wording definitely implies that it's directed at radar controllers.
The radar controller should have coordinated with the Class C tower (if necessary) and Class D tower controllers before switching the pilot to the destination Class D tower controller, if they expected the pilot would have to enter the respective surface areas. The destination Class D tower controller should have coordinated with the other tower controllers, if they expected the pilot to enter those surface areas. If coordination was not effected, they should not issue instructions which would imply permission to enter the other surface areas.
The pilot should query whichever controller they happen to be talking to, and follow the instructions they receive ("proceed through the XXX Class Delta airspace" or "remain outside of XXX Class Delta airspace").

Thank you for providing a detailed answer and for referencing the relevant FAA publication!

Radar services are workload-permitting.  It's still a pilot responsibility to obtain any necessary clearance.  If the pilot is cut loose and told to contact the tower, and an additional Class D exists between where radar service ended, and the destination, it's pilot responsibility to get request and receive clearance through the interim Class D.  It's also his responsibility to avoid airspace with which he has no compliance or clearance.  Receipt of prior service or code does not change that or give an ongoing clearance.


text:
Is axial bearing in turbo fan,turbo jet engines neutraly loaded?
Is axial bearing in turbo fan,turbo jet engines neutraly loaded,if yes then turbine/fan and compresor produce eqaul and opposite force?
If yes ,where then thrust come from,at internal walls,must be some pressure net force to produce thrust.

See: On which point(s) in a jet engine does the reaction force act?

@mins thanks nice explanation

No it generally is not neutral. There is usually some net force. The reason is that very low force is actually bad for bearing life. Look up skidding. So a small imbalance in forces between compressor and turbine is introduced intentionally


text:
Does too big an angle of decalage make the horizontal stabilizer to stall?
This styrofoam airplane has two positions for stabilizer. When put in the upper one, the plane flies in a circle.
Is the stabilizer now stalled because the decalage angle is very big?What happens with the airflow over wing and stabilizer in this configuration?

By circle do you mean a loop, a vertical circle?

Declage controls the AOA to which the static stability forces will be focused with no control input; in other words, the hand's free AOA/speed.  A higher declage angle is the same as increasing elevator input and holding it there.
It changes the trim force balance (nose down pitching moment from the main wing, C of G relative to wing center of pressure, and other appendages that generate pitching moments, opposed by the nose up pitching moment from the tail) to occur at a higher AOA.  Airplanes with moveable horizontal stabilizers, like most jets and a lot of light planes like the Cessna 180/185 family, have what is in effect, adjustable declage as their method of changing trim.  What you are doing by sticking the tail surface in the different slot, they achieve by cranking a screw jack to move the front of the same surface up and down for the same result.
So the tail won't stall at the moderate angle variation there, it'll just drive the main wing to a higher AOA to achieve trim equilibrium, and your glider will trim to a slower speed. If designed properly, if you keep increasing the declage, the main wing will be driven to its stalling AOA, and the aft Center of Pressure shift that occurs at stall will cause the nose down pitching moment to overpower the tail's downward lifting force and the glider pitches over before the tail's airfoil itself stalls.
The result is, when you throw it with the tail stuck in the upper slot, it will pitch up harder for the same throwing velocity as it seeks its higher trim AOA/lower flying speed (if you throw it at exactly its trim speed, it's already in trim when you release it and will just glide straight without pitching).
So bottom line; the lower slot is a "fast slot" and the upper slot is a "slow slot".  The glider will trim to a lower AOA and a higher speed on the lower slot, and vice versa on the slow slot.  That's the practical result of the two options.

With "slow slot" it is almost impossible to fly in straight line ,without adding weight to the nose.It throw it fast it will loop,if you throw it  too slow it will fall to the nose and if you throw it with trim speed for "slow slot it will fly only 2 m. Slow slot use for looping,glider behave like boomerang,always come to you,make circle of about 2m radius

It's looping because it's going way faster than trim speed and it's seeking to slow down.

Can you post all links where is explained why pitch up happens when exceed trim speed?

https://aviation.stackexchange.com/q/72253/42636

The decalage angle of the upper horizontal stablizer slot was measured as arcsin × (.4/2.0 cm) = 11.5 degrees. This symmetrical airfoil will have a high coefficient of (downward) lift when thrown, which would immediately increase the angle of attack of the wing (raise the nose).
11.5 degrees Angle of Attack may be close to stalling, but even if it were stalled it would continue to generate lift.
The only way it could remain stalled would be if the weight were way too far forward.  The glider would just crash into the ground.
If your Center of Gravity was set slightly forward of the wing center of lift (read instructions for your plane as to the where CG should be), the tail will pitch your wing to its correct angle of attack, and both AOAs will be well below stall AOA.
So, if you want to throw that glider off a hill, use the upper slot, set weight forward, and it will fly at a constant speed until it runs out of altitude.  Do not try to throw it too hard when you launch it.
If you want to throw your glider hard for distance (without a hill),
move the weight back and use the lower slot.  This glider will fly at at variable speed until drag slows it down.
CG (weight) setting is critical in both cases.  Try moving it back and forth until the plane flys the way you like it.  (Some free flight models allow you to move the wing.  This is a similar approach).

This glider dont have option to set/move c.g. but you can add some extra weight at nose if you want play with c.g....Yes if you want that glider fly in straight line with upper stabilizer position,you must add lots of weight at nose.As always this works only for one speed,when you exceed trim speed it start pitch up.

In flight the 11.5 degree angle is effectively shared between the wing and the stabiliser; the wing’s AoA will increase and so the stabiliser’s AoA will be smaller (smaller in magnitude, as it’s negative).

@Lost in Space paper clips work well as CG adjusters.  How about posting a picture of your entire plane (and who makes it)!

I don't think CG shifting is nerded

@Abdullah just as in a hang glider, placement of CG is critical to trim settings.  One may make that stabilizer variable, but with FF its easier to move the weight a bit.


text:
How does the pressure distribution at the blades of a radial turbocharger look like?
Consider a piston engine equipped with a turbocharger.
In the turbine section of the turbocharger, the pressurized exhaust airflow comes from the center (radially) to the blades, then changes direction 90 degrees, leaving the blades in axial direction in turbine. (I think airflow maybe half or last 2/3 of blade travel in axial direction?)
What does the pressure distribution at the radial turbine blade look like?
Does the airflow transfer force to the blades using the same principles as on airfoil/wing, where the airflow accelerated over the upper surface of the wing creates low pressure and decelerate at high pressure side ?

I know in axial turbine(picture below),every blade is like "small wing", operate identical as wing in aerodynamic sense, obviously without tip vortex because of blockage of housing

The image looks like the pressurized exhaust comes from the circumference and flows into the centre on the turbine. The same is true of the early turboject engines that had centrifugal turbines.

@JanHudec: Yes. Better drawing. The turbine works like in a Francis design for water (radial then axial).

Please note that the seemingly obvious conclusion obviously without tip vortex because of blockage of housing, is completely wrong since there are geometrical gaps to minimize the risk of contacts between rotating and static parts of the engine. In thurbomachines, this is often called the tip-clearance vortex or tip-leakage vortex. See https://nasa.github.io/JHU-PIV-data/static/documents/Effects%20of%20tip%20clearance%20and%20operating%20condition_Yuanchao_J_Turbo_2019.pdf for further information.

Simulations of pressure and flow conditions over radial turbine blades can be found in this article:
Design and Optimization of a Radial Turbine to Be Used in a Rankine Cycle Operating with an OTEC System
While the article is focussed on Rankine cycle, the images in the article can be considered a good general presentation of radial turbine flow conditions:

As can be anticipated, the pressure is highest at the base/inlet portion of the turbine, and as the gasses travel towards the exit of the turbine, increasing velocity and curvature of the blades will lead to drop in pressure.